On May 29, 2:23 pm, wrote:

I've been following the interesting debate!

It is only interesting that neither can understand the other, and both
supposed to have PhD’s.

I'm going to snip the rest of this, sorry...

I have no complaints. It is getting boring for me.

As I understand it the math of GR needn't scare anyone away because
the same physics applies to the Lienard-Weichert potential of E&M. Of
course, that might be scary enough already for some but..

The math of GR is scary only if you allow yourself to accept mysticism
as wisdom. shrug

Let me try to make a layman's explanation of Carlip's argument.. is
the idea that the space-time ahead of the Earth, where the Earth has
not yet reached in its orbit, has already been affected by
gravitational force of the sun, and therefore the force is in a
slightly different direction making our orbit stable?

Both agree on this point. However, the central argument lies in the
aberration of gravitational effect.

Professor Carlip’s claim is that only the gravitating mass is subject
to the aberrational effect. Yes, the mathematics does show what you
have pointed out and Professor’s aberration cancels out at least to
the first order. However, Dr. Van Flandern, on the other hand,
demands that if aberration is taken into account, the “gravitated”
mass should also be applied to aberration. In this case, it is making
the effect of the speed of gravity much worse.

Professor Carlip cannot be correct if the principle of relativity is
considered. Therefore, with or without any aberrational effect, Dr.
Van Flandern suggests that the only way to explain why the solar
system has not flown apart as Laplace had suggested if the speed of
gravity is any less than several orders of the speed of light.

Despite Dr. Van Flandern’s absurd LR transform, Dr. Van Flandern’s
conclusion based on the observations is valid for now until new laws
of physics are discovered to prove otherwise. LR transform does not
degenerate into Galilean transform and thus has no chance of being a
valid transformation.

Both are heavily intoxicated in mysticism where according to the
Orwellian education,

** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING

carlip-nospam wrote on Sat, 31 May 2008 01:17:57 +0000:

Tom Van Flandern wrote:

This is silly. The potential at the location of the object (not "its
potential" -- the potential is not a characteristic of the orbiting
object, but of the gravitational source) is never constant.

The potential is a characteristic of the source only in the local time
explicit approximation \phi = \phi(r,t).

Which is not the general case of interactions, of course.

For
Newtonian gravity, the derivative of the potential *in the direction of
the orbiting object* is zero, which is why there is no tangential force.
But at each point in the orbit, the derivative of the potential in the
direction perpendicular to the orbit is nonzero. To call such a
potential "constant" is a word game.

This is a pure nonsense, because you are confounding the Newtonian
potential (which has no field interpretation) with a Lienard-Wiechert
potentials (which has field interpretation and associated gradient field
at each r,t point)

Note that when the orbiting object is at a position (x,y,z), the force
is determined by the gradient of the potential at (x,y,z), at the time
the object is at that location.

Right except that technically one is not really taking a (spatial)
gradient.

I know how to compute its gradient at any position and time. Please
write down its "retarded gradient."

Your computation of 'gradients' were corrected in several cited
references.

This statement demonstrates clearly that you don't understand basic
general relativity. The geodesics of the metric *are* the paths of the
target bodies. Once you know the path -- the position as a function of
time -- you can use any coordinate system you like, and any definition
of acceleration you like. The answer is uniquely determined.

In the field formulation? Sure. In the geometrical (geodesic) one? Sure
it is not because the path is coordinate dependent.

That is one of main reasons that one can speak about a gravitational
force in the field formulation. The force is uniquely determined as
orbits of test bodies are also.


--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org

Hi Juan, a quick question below...

On May 31, 6:13 am, "Juan R." González-Álvarez
wrote:
carlip-nospam wrote on Sat, 31 May 2008 01:17:57 +0000:
...
Your computation of 'gradients' were corrected in several cited
references.

This statement demonstrates clearly that you don't understand basic
general relativity. The geodesics of the metric *are* the paths of the
target bodies. Once you know the path -- the position as a function of
time -- you can use any coordinate system you like, and any definition
of acceleration you like. The answer is uniquely determined.

In the field formulation? Sure. In the geometrical (geodesic) one? Sure
it is not because the path is coordinate dependent.

That is one of main reasons that one can speak about a gravitational
force in the field formulation. The force is uniquely determined as
orbits of test bodies are also.

Where U^i is contravariant 3-velocity, then
the geodesic is defined in GR by an "absolute
derivative" like,

DU^i/ds =0,

and that is the equation of motion in GR.
Juan, do you understand that equation?
Regards
Ken S. Tucker

Ken S. Tucker wrote on Sat, 31 May 2008 09:19:15 -0700:

Hi Juan, a quick question below...

On May 31, 6:13 am, "Juan R." González-�lvarez
wrote:
carlip-nospam wrote on Sat, 31 May 2008 01:17:57 +0000:
...
Your computation of 'gradients' were corrected in several cited
references.

This statement demonstrates clearly that you don't understand basic
general relativity. The geodesics of the metric *are* the paths of
the target bodies. Once you know the path -- the position as a
function of time -- you can use any coordinate system you like, and
any definition of acceleration you like. The answer is uniquely
determined.

In the field formulation? Sure. In the geometrical (geodesic) one? Sure
it is not because the path is coordinate dependent.

That is one of main reasons that one can speak about a gravitational
force in the field formulation. The force is uniquely determined as
orbits of test bodies are also.

Where U^i is contravariant 3-velocity, then the geodesic is defined in
GR by an "absolute derivative" like,

DU^i/ds =0,

and that is the equation of motion in GR. Juan, do you understand that
equation? Regards

I wonder by your insistence on posting about stuff you ignore.

Of course, the equation of motion in the geometrical theory of gravity is
geodesic

*But* i wrote

(\blockquote
That is one of main reasons that one can speak about a gravitational
force in the *field* formulation.
)

I can write the geodesic equation of motion in different ways. I like the
expression (in concise notation)

a = - Gamma v v

and i can write it in other ways (ways you do not know :-))

Your query is really funny when one notices i have been writing geodesic
equations both here and in spf.

But could you write the *field* theoretic equation of motion?

No! Well that is not a surprise :-)


--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org

On May 31, 9:58 am, "Juan R." González-Álvarez
wrote:
Ken S. Tucker wrote on Sat, 31 May 2008 09:19:15 -0700:



Hi Juan, a quick question below...

On May 31, 6:13 am, "Juan R." González-Álvarez
wrote:
carlip-nospam wrote on Sat, 31 May 2008 01:17:57 +0000:
...
Your computation of 'gradients' were corrected in several cited
references.

This statement demonstrates clearly that you don't understand basic
general relativity. The geodesics of the metric *are* the paths of
the target bodies. Once you know the path -- the position as a
function of time -- you can use any coordinate system you like, and
any definition of acceleration you like. The answer is uniquely
determined.

In the field formulation? Sure. In the geometrical (geodesic) one? Sure
it is not because the path is coordinate dependent.

That is one of main reasons that one can speak about a gravitational
force in the field formulation. The force is uniquely determined as
orbits of test bodies are also.

Where U^i is contravariant 3-velocity, then the geodesic is defined in
GR by an "absolute derivative" like,

DU^i/ds =0,

and that is the equation of motion in GR. Juan, do you understand that
equation? Regards

I wonder by your insistence on posting about stuff you ignore.

Of course, the equation of motion in the geometrical theory of gravity is
geodesic

*But* i wrote

(\blockquote
That is one of main reasons that one can speak about a gravitational
force in the *field* formulation.
)

I can write the geodesic equation of motion in different ways. I like the
expression (in concise notation)

a = - Gamma v v

and i can write it in other ways (ways you do not know :-))

Your query is really funny when one notices i have been writing geodesic
equations both here and in spf.

But could you write the *field* theoretic equation of motion?

No! Well that is not a surprise :-)

--
Center for CANONICAL |SCIENCE) http://canonicalscience.org

Ok fine, you Juan, TVF and Eugene appear to
be unable to grasp the GPoR emobodied within,

DU^i/ds =0,

succinctly. If you guy's can understand that,
then tell us how to revise that simple equation.

In Eq.(6) at this cite,
http://physics.trak4.com/modern-spacetime.pdf
absolute motion vanishes, in accord with
International Standards.

You see, a good fella like myself, can easily
define the "state of the art", where GToR is
concerned, and all you need to do is show me
where I'm mistaken, go and do it.
Regards
Ken S. Tucker
PS: Stop wasting my time.

Ken S. Tucker wrote on Sat, 31 May 2008 10:26:58 -0700:

On May 31, 9:58 am, "Juan R." González-�lvarez
wrote:
Ken S. Tucker wrote on Sat, 31 May 2008 09:19:15 -0700:



Hi Juan, a quick question below...

On May 31, 6:13 am, "Juan R." González-�lvarez
wrote:
carlip-nospam wrote on Sat, 31 May 2008 01:17:57 +0000:
...
Your computation of 'gradients' were corrected in several cited
references.

This statement demonstrates clearly that you don't understand
basic general relativity. The geodesics of the metric *are* the
paths of the target bodies. Once you know the path -- the
position as a function of time -- you can use any coordinate
system you like, and any definition of acceleration you like. The
answer is uniquely determined.

In the field formulation? Sure. In the geometrical (geodesic) one?
Sure it is not because the path is coordinate dependent.

That is one of main reasons that one can speak about a gravitational
force in the field formulation. The force is uniquely determined as
orbits of test bodies are also.

Where U^i is contravariant 3-velocity, then the geodesic is defined
in GR by an "absolute derivative" like,

DU^i/ds =0,

and that is the equation of motion in GR. Juan, do you understand
that equation? Regards

I wonder by your insistence on posting about stuff you ignore.

Of course, the equation of motion in the geometrical theory of gravity
is geodesic

*But* i wrote

(\blockquote
That is one of main reasons that one can speak about a gravitational
force in the *field* formulation.
)

I can write the geodesic equation of motion in different ways. I like
the expression (in concise notation)

a = - Gamma v v

and i can write it in other ways (ways you do not know :-))

Your query is really funny when one notices i have been writing
geodesic equations both here and in spf.

But could you write the *field* theoretic equation of motion?

No! Well that is not a surprise :-)

--
Center for CANONICAL |SCIENCE) http://canonicalscience.org

Ok fine, you Juan, TVF and Eugene appear to be unable to grasp the GPoR
emobodied within,

DU^i/ds =0,

succinctly. If you guy's can understand that, then tell us how to revise
that simple equation.

In Eq.(6) at this cite,
http://physics.trak4.com/modern-spacetime.pdf absolute motion vanishes,
in accord with International Standards.

You see, a good fella like myself, can easily define the "state of the
art", where GToR is concerned, and all you need to do is show me where
I'm mistaken, go and do it.
Regards
Ken S. Tucker
PS: Stop wasting my time.

Ok, a) you cannot write the field theoretic equation, b) you cannot
understand how the geodesic equation of motion follows from the field
theoretic one in the geometric limit and c) you cannot understand that
extension of SR and GR means :-)


P.S: Your geometric "modern-spacetime.pdf" is useless but you already
were said that in spf, true?

Take dissipation into account and try again :-)


--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org

Ken S. Tucker wrote on Sat, 31 May 2008 10:26:58 -0700:

In Eq.(6) at this cite,
http://physics.trak4.com/modern-spacetime.pdf absolute motion vanishes,
in accord with International Standards.

You see, a good fella like myself, can easily define the "state of the
art", where GToR is concerned, and all you need to do is show me where
I'm mistaken, go and do it.

Ken, to understand _some_ of the multiple mistakes you are doing
regarding time and space and four spacetime continuum, take a look to
preprint and book i cited in the spf thread "100 years of space and time"

You would learn why the Minkowskian geometric four space approach does
not work.



--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org

"Juan R." González-�lvarez wrote on Sat, 31 May 2008 20:24:04 +0200:


You would learn why the Minkowskian geometric four space approach does
not work.

And why your modern spacetime is really very outdated :-)


--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org

On May 31, 10:44 am, "Juan R." González-Álvarez
wrote:
Ken S. Tucker wrote on Sat, 31 May 2008 10:26:58 -0700:



On May 31, 9:58 am, "Juan R." González-Álvarez
wrote:
Ken S. Tucker wrote on Sat, 31 May 2008 09:19:15 -0700:

Hi Juan, a quick question below...

On May 31, 6:13 am, "Juan R." González-Álvarez
wrote:
carlip-nospam wrote on Sat, 31 May 2008 01:17:57 +0000:
...
Your computation of 'gradients' were corrected in several cited
references.

This statement demonstrates clearly that you don't understand
basic general relativity. The geodesics of the metric *are* the
paths of the target bodies. Once you know the path -- the
position as a function of time -- you can use any coordinate
system you like, and any definition of acceleration you like. The
answer is uniquely determined.

In the field formulation? Sure. In the geometrical (geodesic) one?
Sure it is not because the path is coordinate dependent.

That is one of main reasons that one can speak about a gravitational
force in the field formulation. The force is uniquely determined as
orbits of test bodies are also.

Where U^i is contravariant 3-velocity, then the geodesic is defined
in GR by an "absolute derivative" like,

DU^i/ds =0,

and that is the equation of motion in GR. Juan, do you understand
that equation? Regards

I wonder by your insistence on posting about stuff you ignore.

Of course, the equation of motion in the geometrical theory of gravity
is geodesic

*But* i wrote

(\blockquote
That is one of main reasons that one can speak about a gravitational
force in the *field* formulation.
)

I can write the geodesic equation of motion in different ways. I like
the expression (in concise notation)

a = - Gamma v v

and i can write it in other ways (ways you do not know :-))

Your query is really funny when one notices i have been writing
geodesic equations both here and in spf.

But could you write the *field* theoretic equation of motion?

No! Well that is not a surprise :-)

--
Center for CANONICAL |SCIENCE) http://canonicalscience.org

Ok fine, you Juan, TVF and Eugene appear to be unable to grasp the GPoR
emobodied within,

DU^i/ds =0,

succinctly. If you guy's can understand that, then tell us how to revise
that simple equation.

In Eq.(6) at this cite,
http://physics.trak4.com/modern-spacetime.pdfabsolute motion vanishes,
in accord with International Standards.

You see, a good fella like myself, can easily define the "state of the
art", where GToR is concerned, and all you need to do is show me where
I'm mistaken, go and do it.
Regards
Ken S. Tucker
PS: Stop wasting my time.

Ok, a) you cannot write the field theoretic equation, b) you cannot
understand how the geodesic equation of motion follows from the field
theoretic one in the geometric limit and c) you cannot understand that
extension of SR and GR means :-)

P.S: Your geometric "modern-spacetime.pdf" is useless but you already
were said that in spf, true?

Juan, from the standpoint of objective science,
I've attempted to extend to you TVF and Eugene,
the benefit of doubt, and I can argue, on your
behalf for you, given any plausible argument.
The MST "modern-spacetime.pdf" is the best we
have so far, improvements are welcome sir.

But, at least I can provide good theoretics
in 2 pages, posted!

You boys (Juan, TVF and Eugene) need 200 pages
to make your point, what ever the **** it is,
I've still haven't figured out what your trying
to do except being infactuated with Minkowski's
Group(oo), well I know everything about that
Group, so what. Maybe I should advise you guys
on G(oo)!
Regards
Ken S. Tucker
(Hey I'm cheap at 1/2 the cost).

On May 31, 11:58 am, "Ken S. Tucker" wrote:

You boys (Juan, TVF and Eugene) need 200 pages
to make your point, what ever the **** it is,

Well, after several thousands of posts, you still don’t have a point.
shrug